Is (sin(x)+tan(x))/(sin(x))=1+sec(x) ?

The answer to whether (sin(x)+tan(x))/(sin(x))=1+sec(x) is True

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prove (sin(x)+tan(x))/(sin(x))=1+sec(x)

Solution

prove

\frac{sin ( x ) + tan ( x )}{sin ( x )} = 1 + sec (x)

Solution

True

Solution steps

\frac{sin ( x ) + tan ( x )}{sin ( x )} = 1 + sec (x)

Manipulating left side

\frac{sin ( x ) + tan ( x )}{sin ( x )}

Express with sin, cos

\frac{sin ( x ) + tan ( x )}{sin ( x )}

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( x ) + \frac{s i n ( x )}{c o s ( x )}}{sin ( x )}

Simplify

\frac{sin ( x ) + \frac{s i n ( x )}{c o s ( x )}}{sin ( x )} : \frac{cos ( x ) + 1}{cos ( x )}

\frac{sin ( x ) + \frac{s i n ( x )}{c o s ( x )}}{sin ( x )}

Join

sin (x) + \frac{sin ( x )}{cos ( x )} : \frac{sin ( x ) cos ( x ) + sin ( x )}{cos ( x )}

sin (x) + \frac{sin ( x )}{cos ( x )}

Convert element to fraction:

sin (x) = \frac{sin ( x ) cos ( x )}{cos ( x )}

= \frac{sin ( x ) cos ( x )}{cos ( x )} + \frac{sin ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ( x ) cos ( x ) + sin ( x )}{cos ( x )}

= \frac{\frac{s i n ( x ) c o s ( x ) + s i n ( x )}{c o s ( x )}}{sin ( x )}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{sin ( x ) cos ( x ) + sin ( x )}{cos ( x ) sin ( x )}

Factor out common term

sin (x)

= \frac{sin ( x ) ( cos ( x ) + 1 )}{cos ( x ) sin ( x )}

Cancel the common factor:

sin (x)

= \frac{cos ( x ) + 1}{cos ( x )}

= \frac{cos ( x ) + 1}{cos ( x )}

= \frac{1 + cos ( x )}{cos ( x )}

Rewrite using trig identities

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

\frac{1 + \frac{1}{s e c ( x )}}{\frac{1}{s e c ( x )}}

Simplify

\frac{1 + \frac{1}{s e c ( x )}}{\frac{1}{s e c ( x )}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{( 1 + \frac{1}{s e c ( x )} ) sec ( x )}{1}

Join

1 + \frac{1}{sec ( x )} : \frac{sec ( x ) + 1}{sec ( x )}

1 + \frac{1}{sec ( x )}

Convert element to fraction:

1 = \frac{1 sec ( x )}{sec ( x )}

= \frac{1 \cdot sec ( x )}{sec ( x )} + \frac{1}{sec ( x )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot sec ( x ) + 1}{sec ( x )}

Multiply:

1 \cdot sec (x) = sec (x)

= \frac{sec ( x ) + 1}{sec ( x )}

= \frac{\frac{s e c ( x ) + 1}{s e c ( x )} sec ( x )}{1}

Apply the fraction rule:

\frac{a}{1} = a

= \frac{sec ( x ) + 1}{sec ( x )} sec (x)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( sec ( x ) + 1 ) sec ( x )}{sec ( x )}

Cancel the common factor:

sec (x)

= sec (x) + 1

sec (x) + 1

sec (x) + 1

= 1 + sec (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (sin(x)+tan(x))/(sin(x))=1+sec(x) ?
The answer to whether (sin(x)+tan(x))/(sin(x))=1+sec(x) is True